/* nag_opt_simplex(e04ccc) Example Program
 *
 * Copyright 1996 Numerical Algorithms Group.
 *
 * Mark 4, 1996.
 * Mark 6 revised, 2000.
 * Mark 7 revised, 2001.
 *
 */

#include <nag.h>
#include <math.h>
#include <stdio.h>
#include <nag_stdlib.h>
#include <nage04.h>
#include <nagx02.h>

#ifdef __cplusplus
extern "C" {
#endif
static void funct(Integer n, double *xc, double *fc, Nag_Comm *comm);
static void monit(const Nag_Search_State *st, Nag_Comm *comm);
#ifdef __cplusplus
}
#endif

static void ex1(void);
static void ex2(void);

int main(void)
{
  /* Two examples are called, ex1() which uses the
   * default settings to solve the problem and
   * ex2() which solves the same problem with
   * some optional parameters set by the user.
   */

  Vprintf("e04ccc Example Program Results.\n");
  ex1();
  ex2();
  return EXIT_SUCCESS;
}

static void ex1(void)
{
  double objf;
  double x[2];

  Integer n;

  Vprintf("\ne04ccc example 1: no option setting.\n");

  n = 2;
  /* Set up the starting point */
  x[0] = 0.4;
  x[1] = -0.8;

  e04ccc(n, funct, x, &objf, E04_DEFAULT, NAGCOMM_NULL, NAGERR_DEFAULT);

}


static void funct(Integer n, double *xc, double *objf, Nag_Comm *comm)
{
  *objf = exp(xc[0]) * (xc[0] * 4.0 * (xc[0] + xc[1]) +
                        xc[1] * 2.0 * (xc[1] + 1.0) + 1.0);
}
static void ex2(void)
{

  double objf;
  double x[2];

  Integer n;
  Integer monit_freq;

  Boolean print;

  Nag_Comm comm;
  Nag_E04_Opt options;

  static NagError fail;

  Vprintf("\n\ne04ccc example 2: using option setting.\n");

  n = 2;
  monit_freq = 20;
  e04xxc(&options);
  options.print_fun = monit;
  /* Read remaining option value from file */
  fail.print = TRUE;
  print = TRUE;
  e04xyc("e04ccc", "stdin", &options, print, "stdout", NAGERR_DEFAULT);

  comm.p = (Pointer)&monit_freq;

  /* Starting values */
  x[0] = -1.0;
  x[1] = 1.0;

  e04ccc(n, funct, x, &objf, &options, &comm, &fail);

}

static void monit(const Nag_Search_State *st, Nag_Comm *comm)
{
#define SIM(I,J) sim[((I)-1)*n + (J)-1]

  double *sim;
  Integer i, j;
  Integer n, ncall, iter;
  double fmin, fmax;

  Integer *monit_freq=(Integer *)comm->p;

  fmin = st->fmin;
  fmax = st->fmax;
  sim=st->simplex;
  ncall = st->nf;
  iter = st->iter;
  n = st->n;

  if (iter % *monit_freq == 0)
    {
      Vprintf("\nAfter %1ld iteration and %1ld function calls,"
              " the function value is %10.4e\n", iter, ncall, fmin);
      Vprintf("The simplex is\n");
      for (i = 1; i <= n+1; ++i)
        {
          for (j = 1; j <= n; ++j)
            {
              Vprintf(" %12.4e", SIM(i,j));
            }
          Vprintf("\n");
        }
    }
  if (comm->sol_prt)
    {
      Vprintf("The final solution is\n");
      for (i = 0; i <n; i++)
        Vprintf("%12.4e\n", st->x[i]);
      Vprintf("After %1ld iterations and %1ld function calls the function \n"
              "value at the current solution point is %12.4e.\n", iter, ncall, fmin);
    }
} /* monit */